Question 1

A particle with a charge of 1.60 $\times$ 10^-19 and a mass of 1.67 $\times$ 10^-27 kg is found moving in a uniform magnetic field of a strength 1.50 T. The trajectory of the particle is perpendicular to the magnetic field and is circular with a radius of 1.80 cm. The velocity of the particle is

Accepted Answer

The correct answer is A because the velocity of the particle can be calculated using the formula v = qBr/m, where q is the charge of the particle, B is the strength of the magnetic field, r is the radius of the circular trajectory, and m is the mass of the particle. Plugging in the given values, we get v = (1.60 $\times$ 10^-19 C)(1.50 T)(1.80 $\times$ 10^-2 m)/(1.67 $\times$ 10^-27 kg) = 2.6 $\times$ 10⁶ m/s.

Question 2

A particle with a charge of 1.60 $\times$ 10^-19 and a mass of 1.67 $\times$ 10^-27 kg is found moving in a uniform magnetic field of a strength 1.50 T. The trajectory of the particle is perpendicular to the magnetic field and is circular with a radius of 1.80 cm. The kinetic energy of the particle is

Accepted Answer

The kinetic energy (KE) of a charged particle moving in a magnetic field can be calculated using the formula for the centripetal force required for circular motion, $F = \frac{mv^2}{r}$, where $m$ is the mass of the particle, $v$ is its velocity, and $r$ is the radius of the circular path. This force is provided by the magnetic force on a moving charge, which is $F = qvB$, where $q$ is the charge of the particle and $B$ is the magnetic field strength. Setting these two expressions for force equal to each other gives $\frac{mv^2}{r} = qvB$. Solving for $v$ gives $v = \frac{qBr}{m}$. The kinetic energy can then be calculated using $KE = \frac{1}{2}mv^2$. Substituting the expression for $v$ into the kinetic energy formula gives $KE = \frac{1}{2}m\left(\frac{qBr}{m}\right)^2 = \frac{q^2B^2r^2}{2m}$. Substituting the given values ($q = 1.60 \times 10^{-19} C$, $B = 1.50 T$, $r = 1.80 \times 10^{-2} m$, and $m = 1.67 \times 10^{-27} kg$) into this formula and converting the result to keV (1 eV = $1.60 \times 10^{-19} J$) gives a kinetic energy of approximately 27.5 keV.

Question 3

A particle with a charge of 1.60 $\times$ 10^-19 and a mass of 1.67 $\times$ 10^-27 kg is found moving in a uniform magnetic field of a strength 1.50 T. The trajectory of the particle is perpendicular to the magnetic field and is circular with a radius of 1.80 cm. The cyclotron frequency is

Accepted Answer

B

Question 4

The cyclotron frequency depends on the following parameters except

Accepted Answer

The cyclotron frequency, also known as the gyrofrequency, depends on the charge and mass of the particle and the strength of the magnetic field. It does not depend on the velocity of the particle.

Question 5

Two electrons are moving in a uniform magnetic field. The velocity of the first electron is twice the velocity of the second electron. The electron that has the larger circular radius is

Accepted Answer

The radius of the circular path of a charged particle moving in a magnetic field is directly proportional to its velocity. Since the first electron has twice the velocity of the second, its circular path will have a larger radius.

Question 6

Two electrons are moving in a uniform magnetic field. The velocity of the first electron is twice the velocity of the second electron. The electron that has the longer period of motion around the cyclotron is

Accepted Answer

The period of motion for an electron in a magnetic field is independent of its velocity. It is determined by the mass of the electron and the strength of the magnetic field, both of which are constants in this scenario. Therefore, both electrons will have the same period of motion.

Question 7

The current in a wire is directed along the x axis in a uniform magnetic field that is directed along the negative x axis. The force on the wire is along the

Accepted Answer

The force on a current-carrying wire in a magnetic field is given by the cross product of the current direction and the magnetic field direction. When both are parallel or antiparallel, as in this case, the cross product is zero, meaning there is no force on the wire.

Question 8

A 25-cm length of wire that is carrying a current of 4.0 A is parallel to a uniform magnetic field of 2.5 T. The force on the wire is

Accepted Answer

The force on a current-carrying wire in a magnetic field is given by F = I * L * B * sin(θ). Since the wire is parallel to the magnetic field, θ = 0 degrees, and sin(0) = 0, the force is zero.

Question 9

The current in a wire is directed along the y axis in a uniform magnetic field that is directed along the negative x axis. The force on the wire is along the

Accepted Answer

The direction of the force on a current-carrying wire in a magnetic field can be determined using the right-hand rule. Point your fingers in the direction of the current (y-axis), and curl them towards the direction of the magnetic field (negative x-axis). Your thumb will point in the direction of the force, which is the positive z-axis.

Question 10

The current in a wire is directed along the y axis in a uniform magnetic field. The resulting force on the wire is directed along the negative x axis. The magnetic field is directed along the

Accepted Answer

The direction of the force on a current-carrying wire in a magnetic field is given by the right-hand rule. If the current is along the y-axis and the force is along the negative x-axis, the magnetic field must be directed along the -z axis for the right-hand rule to apply.

Question 11

A 20.0-cm length of wire is aligned along the x axis and carries 5.0 mA of current. The wire is in a region where the uniform magnetic field is given by 1.5(i + 2j) mT. The net force on the wire is

Accepted Answer

The force on a current-carrying wire in a magnetic field is given by F = I(L x B). Here, I = 5.0 mA = 5.0 x 10^-3 A, L = 0.20 m in the i direction, and B = 1.5(i + 2j) x 10^-3 T. The cross product L x B results in a force in the k direction with magnitude 3.0 μN.

Question 12

A 20.0-cm length of wire is aligned along the x axis and carries 5.0 mA of current. The wire is in a region where the uniform magnetic field is given by 1.5(i + 3j + 4k) mT. The net force on the wire is

Accepted Answer

The net force on the wire can be calculated using the formula for the magnetic force on a current-carrying wire: $F = I(L \times B)$, where $I$ is the current, $L$ is the length vector of the wire, and $B$ is the magnetic field vector. Given that the wire is aligned along the x-axis, its length vector can be represented as $L = 0.20\hat{i}$ m (since 20.0 cm = 0.20 m), the current $I = 5.0 \times 10^{-3}$ A (since 5.0 mA = 5.0 x 10^{-3} A), and the magnetic field $B = 1.5\hat{i} + 4.5\hat{j} + 6.0\hat{k}$ mT = $1.5 \times 10^{-3}\hat{i} + 4.5 \times 10^{-3}\hat{j} + 6.0 \times 10^{-3}\hat{k}$ T (since 1 mT = 10^{-3} T). The cross product $L \times B$ gives a vector that is perpendicular to both $L$ and $B$, and its magnitude can be calculated as the determinant of a matrix formed by the unit vectors $\hat{i}, \hat{j}, \hat{k}$ and the components of $L$ and $B$. However, since $L$ has only an $\hat{i}$ component, the cross product simplifies to $L \times B = 0.20(0\hat{i} - 6.0\hat{j} + 4.5\hat{k}) = -1.2\hat{j} + 0.9\hat{k}$. The magnitude of this vector is $\sqrt{(-1.2)^2 + (0.9)^2}$ = $\sqrt{1.44 + 0.81}$ = $\sqrt{2.25}$ = 1.5 m. Therefore, the magnitude of the force is $F = I \times 1.5 = 5.0 \times 10^{-3} \times 1.5 = 7.5 \times 10^{-3}$ N, which corresponds to 7.5 mN.

Question 13

A wire carries 5.00 mA of current in a region where the uniform magnetic field strength is 1.50 mT. If the angle between the wire and the magnetic field is 30°, the net force per unit length on the wire is

Accepted Answer

The force per unit length on a current-carrying wire in a magnetic field is given by $F/L = BIl\sin(\theta)$, where $B$ is the magnetic field strength, $I$ is the current, $l$ is the length of the wire (which cancels out in the force per unit length calculation), and $\theta$ is the angle between the wire and the magnetic field. Substituting the given values: $F/L = (1.50 \times 10^{-3} \, \text{T}) \times (5.00 \times 10^{-3} \, \text{A}) \times \sin(30°) = 3.75 \times 10^{-6} \, \text{N/m}$.

Question 14

A wire carries 5.00 mA of current in a region where the uniform magnetic field strength is 1.50 mT. If the net force per unit length on the wire is 4.5 $\mu$ N/m, the angle between the current and the magnetic field is

Accepted Answer

The magnetic force per unit length on a wire is given by $ F/L = I \cdot B \cdot \sin(\theta) $. Solving for $\theta$ using the given values, $\sin(\theta) = \frac{4.5 \times 10^{-6} \, \text{N/m}}{5.00 \times 10^{-3} \, \text{A} \times 1.50 \times 10^{-3} \, \text{T}}$, we find $\theta \approx 37^\circ$.

Question 15

Two parallel wires separated by 1.5 cm both carry 1.5 A of current. The magnitude of the force per unit length on one of the wires is

Accepted Answer

The answer of Two parallel wires separated by 1.5 cm...

Question 16

A force per unit length of 48 $\mu$ N/m is produced when two wires separated by 1.5 mm both carry the same current. The current the wires carry is

Accepted Answer

The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ is given by the formula $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$, where $d$ is the distance between the wires, $L$ is the length of the wires, and $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$). Given that the force per unit length is $48 \, \text{N/m}$, the distance $d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}$, and that both wires carry the same current ($I_1 = I_2 = I$), we can rearrange the formula to solve for $I$:$$48 = \frac{(4\pi \times 10^{-7}) I^2}{2\pi (1.5 \times 10^{-3})}$$Solving for $I$ gives $I \approx 0.60 \, \text{A}$.

Question 17

The force produced by two parallel wires with currents running in the same direction will be

Accepted Answer

The force between two parallel wires carrying currents in the same direction is attractive due to the magnetic fields they generate interacting with each other.

Question 18

A repulsive force is produced between two parallel wires. The relative directions of the currents in the wires

Accepted Answer

A repulsive force between two parallel wires indicates that the currents in the wires are flowing in opposite directions. According to Ampère's law, parallel currents attract when they flow in the same direction and repel when they flow in opposite directions.

Question 19

Two current-carrying wires produce a repulsive force of magnitude F₀. The current in each wire is doubled, resulting in a force whose magnitude is

Accepted Answer

The force between two parallel current-carrying wires is directly proportional to the product of the currents in the wires. If the current in each wire is doubled, the force becomes (2I)*(2I) = 4I^2, which is four times the original force, hence the correct answer is 4F₀.

Question 20

Two current-carrying wires produce a repulsive force of magnitude F₀ when the wires are separated by a distance of r₀. The distance between the two wires is changed to 2r₀, resulting in a force whose magnitude is

Accepted Answer

The force between two current-carrying wires is inversely proportional to the distance between them, according to the formula F ∝ 1/r. Therefore, if the distance is doubled (r becomes 2r), the force becomes half of its original value (F becomes F/2).

Quiz 30: Charges and Currents in Magnetic Fields

A particle with a charge of 1.60 ×\times× 10-19 and a mass of 1.67 ×\times× 10-27 kg is found moving in a uniform magnetic field of a strength 1.50 T. The trajectory of the particle is perpendicular to the magnetic field and is circular with a radius of 1.80 cm. The velocity of the particle is

A particle with a charge of 1.60 ×\times× 10-19 and a mass of 1.67 ×\times× 10-27 kg is found moving in a uniform magnetic field of a strength 1.50 T. The trajectory of the particle is perpendicular to the magnetic field and is circular with a radius of 1.80 cm. The cyclotron frequency is